12.4: Exponential and normal random variables Exponential density function Given a positive constant k > 0, the exponential density function (with parameter k) is

 ke−kx if x ≥ 0 f(x) = 0 if x < 0

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Expected value of an exponential random variable Let X be a continuous random variable with an exponential density function with parameter k. Integrating by parts with u = kx and dv = e−kxdx so that −1 −kx du = kdx and v = k e , we find

Z ∞ E(X) = xf(x)dx −∞ Z ∞ = kxe−kxdx 0 −kx 1 −kx r = lim [−xe − e ]|0 r→infty k 1 = k

2 Variance of exponential random variables Integrating by parts with u = kx2 and dv = e−kxdx so that −1 −kx du = 2kxdx and v = k e , we have

Z ∞ Z r 2 −kx 2 −kx r −kx x e dx = lim ([−x e ]|0 + 2 xe dx) 0 r→∞ 0 2 −kx 2 −kx 2 −kx r = lim ([−x e − xe − e ]|0) r→∞ k k2 2 = k2

2 2 2 1 1 So, Var(X) = k2 − E(X) = k2 − k2 = k2 .

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Example Exponential random variables (sometimes) give good models for the time to failure of mechanical devices. For example, we might measure the number of miles traveled by a given car before its transmission ceases to function. Suppose that this distribution is governed by the exponential distribution with mean 100, 000. What is the probability that a car’s transmission will fail during its first 50, 000 miles of operation?

4 Solution

Z 50,000 1 −1 x Pr(X ≤ 50, 000) = e 100,000 dx 0 100, 000 −x 100,000 50,000 = −e |0 −1 = 1 − e 2 ≈ .3934693403

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Normal distributions The normal density function with mean µ and standard deviation σ is

1 −1 ( x−µ )2 f(x) = √ e 2 σ σ 2π As suggested, if X has this density, then E(X) = µ and Var(X) = σ2. The standard normal density function is the normal density function with µ = σ = 1. That is,

1 −1 x2 g(x) = √ e 2 2π

6 Taylor expansion for the normal cumulative distribution function

1 −1 x2 Let f(x) = √ e 2 be the standard normal density function and 2π R x let F (x) = −∞ f(t)dt be the standard normal cumulative distribution function. We compute a Taylor series expansion,

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Z 1 −1 x2 G(x) = √ e 2 dx 2π ∞ 1 Z X (−1)n = √ x2ndx n!2n 2π n=0 ∞ 1 X (−1)n = √ x2n+1 n!2n(2n + 1) 2π n=0 1 1 1 1 = √ (x − x3 + x5 − x7 + ··· ) 2π 6 40 336

So F (x) = G(x) + C for some C. As 0 is the expected value, we 1 need 2 = F (0) = G(0) + C = C.

8 Example If the continuous random variable X is normally distributed, what is the probability that it takes on a value of more than a standard deviations above the mean?

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Solution Via a change of variables, we may suppose that X is normally distributed with respect to the standard normal distribution. Let F be the cumulative distribution function for the standard normal distribution.

Z ∞ 1 −1 x2 Pr(X ≥ 1) = √ e 2 dx 1 2π = lim (F (r) − F (1)) r→∞ 1 1 13 15 17 ≈ lim (F (r)) − ( + √ (1 − + ) − ) r→∞ 2 2π 6 40 336 ≈ 0.5 − .34332 = .15668

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